Date | May Example question | Marks available | 3 | Reference code | EXM.2.AHL.TZ0.20 |
Level | Additional Higher Level | Paper | Paper 2 | Time zone | Time zone 0 |
Command term | Deduce | Question number | 20 | Adapted from | N/A |
Question
Let γ=1+i√32.
The matrix A is defined by A = (γ101γ).
Deduce that
Show that γ2=γ−1.
Hence find the value of (1−γ)6.
A3 = –I.
A–1 = I – A.
Markscheme
METHOD 1
as γ is a root of z2−z+1=0 then γ2−γ+1=0 M1R1
∴γ2=γ−1 AG
Note: Award M1 for the use of z2−z+1=0 in any way.
Award R1 for a correct reasoned approach.
METHOD 2
γ2=−1+i√32 M1
γ−1=1+i√32−1=−1+i√32 A1
[2 marks]
METHOD 1
(1−γ)6=(−γ2)6 (M1)
=(γ)12 A1
=(γ3)4 (M1)
=(−1)4
=1 A1
METHOD 2
(1−γ)6
=1−6γ+15γ2−20γ3+15γ4−6γ5+γ6 M1A1
Note: Award M1 for attempt at binomial expansion.
use of any previous result e.g. =1−6γ+15γ2+20−15γ+6γ2+1 M1
=1 A1
Note: As the question uses the word ‘hence’, other methods that do not use previous results are awarded no marks.
[4 marks]
A2 = A – I
⇒ A3 = A2 – A M1A1
= A – I – A A1
= –I AG
Note: Allow other valid methods.
[3 marks]
I = A – A2
A–1 = A–1A – A–1A2 M1A1
⇒ A–1 = I – A AG
Note: Allow other valid methods.
[2 marks]